Question

If \[ \vec{a}=\hat{i}+\hat{j}+\hat{k},\quad \vec{c}=\hat{j}-\hat{k}, \] \[ \vec{a}\times \vec{b}=\vec{c} \] and \[ \vec{a}\cdot \vec{b}=3, \] then \(\vec{b}=\)

• A. \(\dfrac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})\)
• B. \(\dfrac{1}{3}(2\hat{i}+5\hat{j}+2\hat{k})\)
• C. \(\dfrac{1}{3}(2\hat{i}+2\hat{j}+3\hat{k})\)
• D. \(\dfrac{1}{3}(2\hat{i}+5\hat{j}+5\hat{k})\)

Hint:

When an unknown vector is involved in both dot product and cross product conditions, assume \[ \vec{b}=x\hat{i}+y\hat{j}+z\hat{k} \] and compare coefficients after expanding.

Answer 1

The Correct Option is A

Step 1: Assume the unknown vector.
Let \[ \vec{b}=x\hat{i}+y\hat{j}+z\hat{k} \] Given, \[ \vec{a}=\hat{i}+\hat{j}+\hat{k} \]

Step 2: Use the dot product condition.
Given, \[ \vec{a}\cdot \vec{b}=3 \] So, \[ (\hat{i}+\hat{j}+\hat{k})\cdot (x\hat{i}+y\hat{j}+z\hat{k})=3 \] \[ x+y+z=3 \]

Step 3: Compute \(\vec{a}\times \vec{b}\).
\[ \vec{a}\times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & 1 & 1\\ x & y & z \end{vmatrix} \] \[ = \hat{i}(z-y)-\hat{j}(z-x)+\hat{k}(y-x) \] \[ = (z-y)\hat{i}+(x-z)\hat{j}+(y-x)\hat{k} \]

Step 4: Compare with \(\vec{c}\).
Given, \[ \vec{c}=\hat{j}-\hat{k} \] So, \[ \vec{a}\times \vec{b}=\vec{c} \] \[ (z-y)\hat{i}+(x-z)\hat{j}+(y-x)\hat{k} = 0\hat{i}+1\hat{j}-1\hat{k} \] Comparing coefficients, \[ z-y=0 \] \[ x-z=1 \] \[ y-x=-1 \]

Step 5: Solve the equations.
From \[ z-y=0, \] we get \[ z=y \] From \[ x-z=1, \] we get \[ x=y+1 \] Now use \[ x+y+z=3 \] Substituting \(z=y\) and \(x=y+1\), \[ (y+1)+y+y=3 \] \[ 3y+1=3 \] \[ 3y=2 \] \[ y=\frac{2}{3} \]

Step 6: Find \(x\) and \(z\).
Since \[ z=y, \] we get \[ z=\frac{2}{3} \] Also, \[ x=y+1 \] \[ x=\frac{2}{3}+1=\frac{5}{3} \] Therefore, \[ \vec{b} = \frac{5}{3}\hat{i}+\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} \] \[ = \frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k}) \]

Step 7: Final conclusion.
Hence, \[ \boxed{\frac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})} \]